Problem: A secant line intersects the graph of $f(x)=-3x^2+1$ at two points with $x$ -coordinates $3$ and $3+h$, where $h\neq0$. What is the slope of the secant line in terms of $h$ ? Your answer must be fully expanded and simplified.
Explanation: We are given that the secant line intersects the graph at $x=3$ and $x=3+h$. Since these points are on the the graph of $f(x)=-3x^2+1$, we know that they must be $(3,-26)$ and $(3+h,\,-3(3+h)^2+1)$, respectively. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{-3(3+h)^2+1-(-26)}{3+h-3} \\\\ &=\dfrac{-3(3+h)^2+27}{h} \end{aligned}$ We can now simplify the expression we obtained. $\begin{aligned} &\phantom{=}\dfrac{-3(3+h)^2+27}{h} \\\\ &=\dfrac{-3(9+6h+h^2)+27}{h} \\\\ &=\dfrac{-27-18h-3h^2+27}{h} \\\\ &=\dfrac{-3h^2-18h}{h} \\\\ &=\dfrac{h(-3h-18)}{h} \\\\ &=-3h-18\text{, for }h\neq 0 \end{aligned}$ Since we are given that $h\neq 0$, we can conclude that the slope of the secant line is $-3h-18$.